Does $\sum_{n=1}^\infty n^{-1-|\sin n|^a}$ converge for some $a\in(0,1)$?

Question

The divergence of the series $\sum_{n=1}^\infty n^{-1-|\sin n|}$ is proved here. An inmediate consequence is that if $a\ge1$ then $\sum_{n=1}^\infty n^{-1-|\sin n|^a}$ also diverges. My question is: is there some $a\in(0,1)$ such that $$ \sum_{n=1}^\infty \frac{1}{n^{1+|\sin n|^a}}<\infty? $$ I have tried to adapt the proof given in the above link, but I have been unable to do it. On the other hand, it can be adapted to prove that $$ \sum_{n=1}^\infty \frac{1}{n^{1+b|\sin n|}}=\infty\quad\forall b>0. $$

Answer 1

As Greg Martin suggests, the convergence of this series for all $0<a<1$ can be established by a variation of my answer for the related question.

Recall from that answer that known bounds on the irrationality measure for $\pi$ lead to effective bounds on the discrepancy of the sequence $s_n := \{\frac n\pi \bmod 1\}$, which determines the value of $\lvert\sin n\rvert$. More precisely, there is an absolute constant $c>0$ such that, uniformly for $0 \le \alpha \le \beta \le 1,$ we have the equidistribution estimate

$$\# \{ n \le N : s_n \in [\alpha,\beta] \} = (\beta-\alpha)N + O(N^{1-c}).\tag{1}$$

(The constant factor implied by the $O(N^{1-c})$ may be taken as $1$ if we choose $c$ small enough.)

In the other question, a pigeonhole argument sufficed to get lower bounds on this counting function, but to prove convergence we need upper bounds. The point of $(1)$ is that the values of $\sin n$ for $n \le N$ are distributed like uniformly random values of $\sin x$, provided that our histogram's buckets are not so narrow as a function of $N$.

We need to be a bit careful here as we'll want to apply $(1)$ with a slowly-shrinking interval. So let's partition $\mathbb N$ into dyadic blocks $[2^k,2^{k+1})$ and consider the sum over each block. Note that $(1)$ is insensitive to offsets: if we consider $s_{n+m}$ rather than $s_n$, the same estimate holds because we are just adding a constant (modulo 1) to each term, which just shifts the interval $[\alpha,\beta]$ along the torus (wraparound effects only adjust the constant and can be subsumed into $c$).

Fix any $A$ such that $1 < A < a^{-1}$ (this is possible since $a<1$). For each $k\ge 1$, divide the block $[2^k,2^{k+1})$ into $S_k$ and $T_k$ where $$S_k := \{n \in [2^k,2^{k+1}) : \lvert \sin n \rvert < k^{-A} \}$$ consists of indices where $\lvert \sin n\rvert$ is very small, and $T_k$ consists of all the rest.

Since the dyadic block has length $2^k$ and this is larger than any power of $k^A$, we can deduce from $(1)$ that $\lvert S_k \rvert \le C\cdot k^{-A} 2^k$ except for perhaps finitely many values of $k$ (depending only on $A$), which we can safely ignore.

For $n \in S_k$, we crudely bound $n^{-1-\lvert\sin n\rvert^a}$ above by $1/n \le 1/2^k$: the sum of our series taken over $S_k$ for a fixed $k$ is at most $C k^{-A}$. Summing this over all blocks $S_k$, this converges since $A>1$.

For $n \in T_k$, we have

$$ \frac{1}{n^{1+\lvert \sin n\rvert^a}} \le \frac1n n^{-k^{-Aa}} \le 2^{-k} 2^{k(-k^{-Aa})} = 2^{-k} 2^{-k^{1-Aa}}.$$

Since crudely $\lvert T_k \rvert \le 2^k$, the sum over $T_k$ is at most $2^{-k^{1-Aa}}$. Since $Aa < 1$, the uppermost exponent is positive, so when we sum this bound over $k$ it also converges (for large $k$ it compares favourably with, say, $1/k^2$).